Two point charges \( q_1=4\,\mu\mathrm{C} \) and \( q_2=9\,\mu\mathrm{C} \) are placed 20 cm apart. The electric field due to them will be zero on the line joining them at a distance of
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8 cm from \( q_1 \)
Both charges are positive, so on the line between them their fields point in opposite directions. The field is zero at the point between them where the two magnitudes \( kq/r^{2} \) are equal; this point lies nearer the smaller charge \( q_1 \).
Let the point be at a distance \( x \) from \( q_1 \), so it is \( 20\,\mathrm{cm}-x \) from \( q_2 \). Equating the magnitudes of the two fields:
\( \displaystyle \frac{k\,(4\,\mu\mathrm{C})}{x^{2}}=\frac{k\,(9\,\mu\mathrm{C})}{(20\,\mathrm{cm}-x)^{2}} \)
\( \displaystyle \left(\frac{20\,\mathrm{cm}-x}{x}\right)^{2}=\frac{9}{4} \)
Taking the positive root, since the point lies between the charges:
\( \displaystyle \frac{20\,\mathrm{cm}-x}{x}=\frac{3}{2} \)
\( \Rightarrow\; 40\,\mathrm{cm}-2x=3x \)
\( x=8\,\mathrm{cm} \)
Therefore, the field is zero 8 cm from \( q_1 \), that is, 12 cm from \( q_2 \).